/*
 * @lc app=leetcode.cn id=200 lang=cpp
 *
 * [200] 岛屿数量
 *
 * https://leetcode.cn/problems/number-of-islands/description/
 *
 * algorithms
 * Medium (58.50%)
 * Likes:    1878
 * Dislikes: 0
 * Total Accepted:    544.9K
 * Total Submissions: 930.6K
 * Testcase Example:  '[["1","1","1","1","0"],["1","1","0","1","0"],["1","1","0","0","0"],["0","0","0","0","0"]]'
 *
 * 给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。
 * 
 * 岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
 * 
 * 此外，你可以假设该网格的四条边均被水包围。
 * 
 * 
 * 
 * 示例 1：
 * 
 * 
 * 输入：grid = [
 * ⁠ ["1","1","1","1","0"],
 * ⁠ ["1","1","0","1","0"],
 * ⁠ ["1","1","0","0","0"],
 * ⁠ ["0","0","0","0","0"]
 * ]
 * 输出：1
 * 
 * 
 * 示例 2：
 * 
 * 
 * 输入：grid = [
 * ⁠ ["1","1","0","0","0"],
 * ⁠ ["1","1","0","0","0"],
 * ⁠ ["0","0","1","0","0"],
 * ⁠ ["0","0","0","1","1"]
 * ]
 * 输出：3
 * 
 * 
 * 
 * 
 * 提示：
 * 
 * 
 * m == grid.length
 * n == grid[i].length
 * 1 
 * grid[i][j] 的值为 '0' 或 '1'
 * 
 * 
 */

// @lc code=start
class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        int res = 0;
        int m = grid.size();
        if (m == 0) {
            return res;
        }
        int n = grid[0].size();
        
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == '1') {
                    ++res;
                    grid[i][j] = '0';
                    queue<pair<int, int>> neighbors;
                    neighbors.emplace(i, j);
                    while (!neighbors.empty()) {
                        auto [row, col] = neighbors.front();
                        neighbors.pop();
                        if (row - 1 >= 0 && grid[row - 1][col] == '1') {
                            neighbors.emplace(row - 1, col);
                            grid[row - 1][col] = '0';
                        }
                        if (row + 1 < m && grid[row + 1][col] == '1') {
                            neighbors.emplace(row + 1, col);
                            grid[row + 1][col] = '0';
                        }
                        if (col - 1 >= 0 && grid[row][col - 1] == '1') {
                            neighbors.emplace(row, col - 1);
                            grid[row][col - 1] = '0';
                        }
                        if (col + 1 < n && grid[row][col + 1] == '1') {
                            neighbors.emplace(row, col + 1);
                            grid[row][col + 1] = '0';
                        }
                    }
                }
            }
        }

        return res;       
    }
};
// @lc code=end

